Exercise 5.2.1: Magnetic field of a charging capacitor

The charge on the plate is $q$, so $D=q/(\pi R^2)$ in the gap and $J_d=\partial D/\partial t=i/(\pi R^2)$, uniform for $\rho <R$ and zero outside. Apply Eq. (5.2.5) to a circle of radius $\rho$ centred on the axis ($J=0$ in the gap):

With $i=0.5\mathrm{A}$ and $R=0.05\mathrm{m}$: $H_{\varphi }(2\mathrm{cm})=0.5\times 0.02/(2\pi \times 0.0025)=0.637A/m$, at the rim $0.5/(2\pi \times 0.05)=1.59A/m$, and at $10\mathrm{cm}$ $0.5/(2\pi \times 0.1)=0.796A/m$.

Outside the plates the field is $i/(2\pi \rho )$, exactly the field of the feeding wire: the magnetic field does not notice the gap. Inside, the field grows linearly with $\rho$, like the field inside a wire carrying a uniform current, with the displacement current playing the role of that current (Figure 5.2.3b).

Both expressions give $i/(2\pi R)$ at $\rho =R$. $B={\mu }_{0}H$ at the rim is $4\pi \times 10^{-7}\times 1.59=2.0\mu \mathrm{T}$, about one twenty-fifth of the Earth's field of roughly $50\mu \mathrm{T}$: this is why the displacement current was deduced from the equations before it was measured.